关闭弹出窗口后重新加载父页面 [英] Reload parent page after closing popup window
问题描述
我正在尝试让用户通过弹出窗口登录.当他们单击指向作为php变量的弹出窗口的链接时，他们可以登录.当窗口关闭时，我希望它重新加载他们最初位于的页面(父页面).
I am trying to have the user sign in through a popup window. When they click the link to the popup window which is a php variable they can sign in. When the window closes I want it to reload the page they were originally on (the parent page).
这是signin.php页面上的代码...
Here is the code on the signin.php page...
<body onunload="opener.location=('')">
但这只是使登录页面成为用户所在的页面.我想我需要在括号中加上一些内容，但我不知道该怎么办.
But all this does is make the sign in page become the page the user was on. I think I need to put something in the parentheses, but I can't figure out what goes there.
推荐答案
要重新加载页面，可以将location属性设置为当前值，如下所示:
To reload a page, you can set the location property as the current value, like this:
window.location = window.location;
因此，对于您的情况，应使用字面意思:
So for your case, you would use, literally:
onunload="window.opener.location = window.opener.location;"
您还可以使用location对象的reload方法:
You can also use the reload method of the location object:
onunload="window.opener.location.reload();"
这是首选方法.
此外，请参考上一个问题的已接受答案:刷新后的父窗口关闭弹出窗口
Also, please refer to the accepted answer for your previous question: Refreshing Parent window after closing popup
文档
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MDN上的
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window.location- https://developer.mozilla.org/en /DOM/window.location
MDN上的 -
window.opener- https://developer.mozilla.org /Talk:zh/DOM/window.opener
window.locationon MDN - https://developer.mozilla.org/en/DOM/window.locationwindow.openeron MDN - https://developer.mozilla.org/Talk:en/DOM/window.opener
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